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BITSAT 2014
Exam
BITSAT
Year
2014
Subject
Chemistry
Chapter
Thermodynamics (C)

BITSAT 2014 · Chemistry · Thermodynamics (C)

If the enthalpy of vaporization of water is 186.5 kJmol^ -1 , the...

If the enthalpy of vaporization of water is 186.5 kJmol−1\mathrm{kJmol}^{-1}, the entropy if its vaporization will be:

Options

  1. A0.5kJK−1mol−10.5 \mathrm{kJK}^{-1} \mathrm{ mol}^{-1}
  2. B1.0kJK−1mol−11.0 \mathrm{kJK}^{-1} \mathrm{ mol}^{-1}
  3. C1.5kJK−1mol−11.5 \mathrm{kJK}^{-1} \mathrm{ mol}^{-1}
  4. D2.0kJK−1mol−12.0 \mathrm{kJK}^{-1} \mathrm{ mol}^{-1}

Correct answer

A. 0.5kJK−1mol−10.5 \mathrm{kJK}^{-1} \mathrm{ mol}^{-1}

Step-by-step solution

Given enthalpy of vaporization, ΔH=186.5kJmol−1\Delta \mathrm{H}=186.5 \mathrm{ kJ} \mathrm{ mol}^{-1} Boiling point of water =100^{{}^{\circ}} C=100+273=373K\mathrm{C}=100+273=373 \mathrm{ K} Entropy change, ΔS=ΔHT=186.5kJmol−1373K=0.5kJmol−1K−1\begin{array}{r} \Delta \mathrm{S}=\frac{\Delta \mathrm{H}}{\mathrm{T}}=\frac{186.5 \mathrm{ kJ} \mathrm{ mol}^{-1}}{373 \mathrm{ K}}\\ =0.5 \mathrm{ kJ} \mathrm{ mol}^{-1} \mathrm{ K}^{-1} \end{array}
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